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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 242959 Accepted Submission(s):
46863
Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
注意给数组清零
#include<stdio.h> #include<stdlib.h> #include<string.h> int main() { int m,j,n,i,l1,l2,t,k=1; char s1[1100]; char s2[1100]; int s3[1100]; int s4[1100]; scanf("%d",&n); while(n--) { scanf("%s %s",s1,s2); memset(s3,0,sizeof(s3)); //给数组清0 memset(s4,0,sizeof(s4)); l1=strlen(s1); l2=strlen(s2); t=0; for(j=l1-1,i=0;j>=0;j--,i++) s3[i]=s1[j]-‘0‘; //将字符串转换为数字 for(j=l2-1,i=0;j>=0;i++,j--) s4[i]=s2[j]-‘0‘; for(i=0;i<1100;i++) { s3[i]+=s4[i]; if(s3[i]>=10) //对各位求和 { s3[i]-=10; s3[i+1]++; } } printf("Case %d:\n",k++); printf("%s + %s = ",s1,s2); for(i=1099;i>=0;i--) if(s3[i]!=0) break; for(;i>=0;i--) printf("%d",s3[i]); printf("\n"); if(n>0) printf("\n"); } return 0; }
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原文地址:http://www.cnblogs.com/tonghao/p/4356781.html